Re: My reviewers think that halt deciders must report on the behavior of their caller

Op 06.jul.2025 om 17:36 schreef olcott:

On 7/6/2025 4:24 AM, Fred. Zwarts wrote:

Op 05.jul.2025 om 18:12 schreef olcott:

On 7/5/2025 2:39 AM, Fred. Zwarts wrote:

Op 05.jul.2025 om 00:15 schreef olcott:

On 7/4/2025 3:53 PM, Richard Damon wrote:

On 7/4/25 4:43 PM, olcott wrote:

On 6/3/2025 10:02 PM, dbush wrote:

On 6/3/2025 10:58 PM, olcott wrote:

On 6/3/2025 9:46 PM, dbush wrote:

On 6/3/2025 10:34 PM, olcott wrote:

On 6/3/2025 9:12 PM, dbush wrote:

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Given any algorithm (i.e. a fixed immutable sequence of instructions) X described as <X> with input Y:
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A solution to the halting problem is an algorithm H that computes the following mapping:
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(<X>,Y) maps to 1 if and only if X(Y) halts when executed directly
(<X>,Y) maps to 0 if and only if X(Y) does not halt when executed directly
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Yes there is no algorithm that does that

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Excellent!
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Let The Record Show
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That Peter Olcott
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Has *EXPLICITLY* admitted
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That no algorithm H exists that meets the above requirements, which is precisely the theorem that the halting problem proofs prove.

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In the exact same way that there is no set of all set
that contain themselves. ZFC did not solve Russell's
Paradox as much as it showed that Russell's Paradox
was anchored in an incoherent foundation, now called
naive set theory.

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Which arose because the axioms of naive set theory created a contradiction.
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Likewise with halt deciders that are required to report
on the behavior of directly executed Turing machines.

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And what is the CONTRADICTION?
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The result is just some things are not computable.
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Directly executed Turing machines are outside of the
domain of every Turing machine decider.

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Then so is mathematics, as "numbers" can't be given to Turing Machines, only representations of them.
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Numbers always work the same way so it makes no difference.
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*HHH(DDD)==0 and HHH1(DDD)==1 are both correct*
https://claude.ai/share/da9b8e3f-eb16-42ca-a9e8-913f4b88202c
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When we compare DDD emulated by HHH and DDD emulated
by HHH1 SIDE-BY-SIDE. (Mike didn't do it this way).
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*The difference is when*
HHH begins to simulate itself simulating DDD and
HHH1 NEVER begins to simulate itself simulating DDD.
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Misleading words. Both simulators are given the same input. Both are simulating the same input, which includes HHH, not HHH1. So, there is no difference.

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*Its just over your head*
HHH(DDD) Does simulate itself simulating DDD
thus causing recursive simulation that cannot
possibly reach the simulated final halt state of DDD.
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It is apparently over your head that when HHH is unable to reach the end of the simulation, this is not a proof that the end does not exist.

 Something like mathematical induction proves that
when an infinite number of steps of DDD are simulated
by HHH according to the semantics of the C programming
language that this simulated DDD does not reach its
own simulated "statement. All of the AI bots figure
this out on their own.

Only if you cheat with the induction, by also changing the input.
Even then the induction only proves that, no matter how many steps are simulated, HHH always fails to reach the abort code specified in the input, so, it fails to reach the end.
Mathematical induction', therefore, only proves the failure of HHH.
When we limit ourselves to the input presented in your code example, we see that many simulators are able to reach the final halt state in a finite number of steps, but HHH gives up before it has simulated the same number of steps.
There is nothing wrong with giving up a simulation when there is suspicion that it could be an infinite simulation. But in that case, simulation fails as analysis of the behaviour and another method is needed to analyse the behaviour.